132 (2007)
MATHEMATICA BOHEMICA
No. 2, 185–203
ON γ-LABELINGS OF ORIENTED GRAPHS
Futaba Okamoto, La Crosse, Ping Zhang, Kalamazoo,
Varaporn Saenpholphat, Bangkok
(Received January 24, 2006)
Abstract. Let D be an oriented graph of order n and size m. A γ-labeling of D is
a one-to-one function f : V (D) → {0, 1, 2, . . . , m} that induces a labeling f 0 : E(D) →
{±1, ±2, . . . , ±m} of the arcs of D defined byPf 0 (e) = f (v) − f (u) for each arc e = (u, v) of
D. The value of a γ-labeling f is val(f ) =
f 0 (e). A γ-labeling of D is balanced if the
e∈E(G)
value of f is 0. An oriented graph D is balanced if D has a balanced labeling. A graph G
is orientably balanced if G has a balanced orientation. It is shown that a connected graph
G of order n > 2 is orientably balanced unless G is a tree, n ≡ 2 (mod 4), and every vertex
of G has odd degree.
Keywords: oriented graph, γ-labeling, balanced γ-labeling, balanced oriented graph,
orientably balanced graph
MSC 2000 : 05C20, 05C78
1. Introduction
Let G be a graph of order n and size m. A γ-labeling of G is defined in [1] as a
one-to-one function f : V (G) → {0, 1, 2, . . . , m} that induces a labeling f 0 : E(G) →
{1, 2, . . . , m} of the edges of G defined by f 0 (e) = |f (u) − f (v)| for each edge e = uv
of G. Therefore, a graph G of order n and size m has a γ-labeling if and only if
m > n − 1. Each γ-labeling f is assigned a value denoted by val(f ) and defined by
X
val(f ) =
f 0 (e).
e∈E(G)
Since f is a one-to-one function from V (G) to {0, 1, 2, . . . , m}, it follows that f 0 (e) > 1
for each edge e in G and so val(f ) > m.
Research supported in part by the Thailand Research Fund MRG4780208.
185
Figure 1 shows nine γ-labelings f1 , f2 , . . . , f9 of the path P5 of order 5 (where the
vertex labels are shown above each vertex and the induced edge labels are shown
below each edge). The value of each γ-labeling is shown in Figure 1 as well.
f1 :
f4 :
f7 :
0
1
1
3
2 3 4
1 1 1
val(f1 ) = 4
2
1
4
0 1 4
2 1 3
val(f4 ) = 7
0
4
3 1 2
3 2 1
val(f7 ) = 10
f2 :
f5 :
f8 :
0
1
1
2
2 4 3
1 2 1
val(f2 ) = 5
1
1
2
3 4 0
2 1 4
val(f5 ) = 8
0
3
3 1 4
3 2 3
val(f8 ) = 10
f3 :
f6 :
f9 :
2
3
4 1 0
1 3 1
val(f3 ) = 6
1
0
3
2 4 1
1 2 3
val(f6 ) = 9
3
3
0
3
4 1 2
4 3 1
val(f9 ) = 11
Figure 1: Some γ-labelings of P5
For a graph G of order n and size m, the maximum value valmax (G) of G is defined
as the maximum value among all γ-labelings of G and the minimum value valmin (G)
is the minimum value among all γ-labelings of G, that is,
valmax (G) = max{val(f ) : f is a γ-labeling of G},
valmin (G) = min{val(f ) : f is a γ-labeling of G}.
It turns out that valmax (P5 ) = 11 and valmin (P5 ) = 4.
If the induced edge-labeling f 0 of a γ-labeling f of a graph is also one-to-one,
then f is a graceful labeling. Among all labelings of graphs, graceful labelings are
probably the best known and most studied. Graceful labelings originated with a
1967 paper of Rosa [6], who used the term β-valuations. A few years later, Golomb
[5] called these labelings “graceful” and this is the terminology that has been used
since then. Gallian [4] has written an extensive survey on labelings of graphs. The
subject of γ-labelings of graphs was studied in [1], [2]. In this paper, we study γlabelings of oriented graphs. We refer to the book [3] for graph-theoretical notation
and terminology not described in this paper.
If a digraph D has the property that for each pair u, v of distinct vertices of
D, at most one of (u, v) and (v, u) is an arc of D, then D is an oriented graph.
Thus an oriented graph D is obtained by assigning a direction to each edge of some
graph G. In this case, the digraph D is also called an orientation of G. For an
oriented graph D of order n and size m, a γ-labeling of D is a one-to-one function
f : V (D) → {0, 1, 2, . . . , m} that induces a labeling
f 0 : E(D) → {±1, ±2, . . . , ±m}
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of the arcs of D defined by f 0 (e) = f (v) − f (u) for each arc e = (u, v) of D. The
value of a γ-labeling f is denoted by val(f : D) and is defined by
X
val(f : D) =
f 0 (e).
e∈E(D)
If the oriented graph D under consideration is clear, we use val(f ) to denote
val (f : D). For example, γ-labelings of two oriented graphs are shown in Figure 2,
where val(f1 ) = 2 and val(f2 ) = 0.
f1 : 0
1
1
2
3
−1
f2 : 0
2
−1
val(f1 ) = 2
1
2
3
−1
2
val(f2 ) = 0
Figure 2: Illustrating γ-labelings of digraphs
For a digraph D, the converse D ∗ of D is the digraph obtained from D by reversing
the direction of every arc of D. If f is a γ-labeling of D, then f is also a γ-labeling
of D∗ and
(1)
val(f, D∗ ) = − val(f, D).
The following lemma will be useful.
Lemma 1.1. Let G be a nonempty graph that is decomposed into graphs
G1 , G2 , . . . , Gk (k > 2). For 1 6 i 6 k, let Di be an orientation of Gi and let D be
k
P
the orientation of G for which E(D) =
E(Di ). If f is a γ-labeling of D and fi is
i=1
the restriction of f to Di for 1 6 i 6 k, then
val(f : D) =
k
X
val(fi : Di ).
i=1
187
2. Balanced γ-labelings
A γ-labeling f of an oriented graph D is balanced if the value of f is 0 and an
oriented graph having a balanced labeling is called a balanced digraph. For example,
the γ-labeling f1 in Figure 2 is not balanced, while the γ-labeling f2 in Figure 2 is
balanced and so the second digraph in Figure 2 is a balanced digraph. In fact, the
first digraph of Figure 2 is not a balanced digraph. In general, if P~n : v1 , v2 , . . . , vn
is a nontrivial directed path of order n and f is a γ-labeling of P~n , then
val(f ) =
X
0
f (e) =
n−1
X
[f (vi+1 ) − f (vi )] = f (vn ) − f (v1 ) 6= 0.
i=1
~n )
e∈E(P
This gives the following observation.
Observation 2.1. No γ-labeling of a nontrivial directed path is balanced and so
no nontrivial directed path is balanced.
~ n : v1 , v2 , . . . , vn , vn+1 = v1 be a directed cycle of order
On the other hand, let C
~ n . Then
n > 3 and let f be a γ-labeling of C
val(f ) =
X
f 0 (e)
~n)
e∈E(C
=
n
X
[f (vi+1 ) − f (vi )] = f (vn+1 ) − f (v1 ) = 0.
i=1
This gives the following observation.
Observation 2.2. Every γ-labeling of a directed cycle is balanced and every
directed cycle is balanced.
By Lemma 1.1 and Observations 2.1 and 2.2, we have the following.
Proposition 2.3. Let f be a γ-labeling of an oriented graph D.
(a) If D can be decomposed into k directed ui − vi paths for 1 6 i 6 k, then
val(f ) =
k
X
[f (vi ) − f (ui )].
i=1
(b) If D can be decomposed into directed cycles, then f (and therefore D) is balanced.
188
An oriented graph D is Eulerian if D possesses a directed circuit that contains
all of the arcs and vertices of D. Furthermore, an oriented graph D is Eulerian if
and only if D can be decomposed into directed cycles. The following is therefore an
immediate consequence of Proposition 2.3(b).
Corollary 2.4. Every Eulerian oriented graph is balanced.
Indeed, every γ-labeling of an Eulerian oriented graph is balanced by Observation 2.2. In fact, Eulerian oriented graphs are the only oriented graphs having this
property. In order to show this, we first state a lemma. For a vertex v in an oriented
graph, the number of vertices to which a vertex v is adjacent is the outdegree of v and
is denoted by od v. The number of vertices from which v is adjacent is the indegree
of v and is denoted by id v.
Lemma 2.5. If f is a γ-labeling of an oriented graph D, then
(2)
val(f ) =
X
(id v − od v)f (v).
v∈V (D)
.
For each vertex v, the term f (v) appears exactly id v times in the sum
of val(f ); while the term −f (v) appears exactly od v times in val(f ). Therefore, (2)
holds.
Theorem 2.6. Let D be an oriented graph. Then every γ-labeling of D is
balanced if and only if D is Eulerian.
.
We have seen that every γ-labeling of an Eulerian oriented graph is
balanced. Thus it remains to verify the converse. Assume that D is not Eulerian.
We show that there is a γ-labeling of D that is not balanced. Let f be a γ-labeling
of D. If val(f ) 6= 0, then we have the desired result. Thus we may assume that
val(f ) = 0. Let
U = {v ∈ V (D) : id v > od v}
and
W = {v ∈ V (D) : id v < od v}.
Since D is not Eulerian, U 6= ∅ and W 6= ∅. Hence there exist some vertex u ∈ U
and some vertex w ∈ W . We define a γ-labeling g of D by

f (w) if v = u,


g(v) = f (u) if v = w,


f (v) if v =
6 u, w.
189
Suppose that f (w) − f (u) = c 6= 0. It then follows by (2) that
val(g) =
X
(id v − od v)g(v)
v∈V (D)
=
X
(id v − od v)g(v)
v∈V (D)−{u,w}
+ (id u − od u)g(u) + (id w − od w)g(w)
X
(id v − od v)f (v)
=
v∈V (D)−{u,w}
+ (id u − od u)f (w) + (id w − od w)f (u)
X
(id v − od v)f (v)
=
v∈V (D)
+ (id u − od u)(f (w) − f (u)) + (id w − od w)(f (u) − f (w))
= c[(id u − od u) + (od w − id w)] 6= 0.
Thus g is not balanced, as claimed.
3. Orientably balanced graphs
A graph G is orientably balanced if G has a balanced orientation. By Observation 2.2, every cycle is orientably balanced. In fact, more can be said. A graph G
is Eulerian if G possesses a circuit containing all of the edges and vertices of G. In
particular, every cycle is Eulerian. It is well-known that a graph G has an Eulerian
orientation if and only if G is Eulerian. Thus, the following is a consequence of
Theorem 2.6.
Proposition 3.1. Every Eulerian graph is orientably balanced.
Certainly, every γ-labeling of an orientation of the path P2 has value 1 or −1 (see
Figure 3) and so P2 is not orientably balanced. On the other hand, both P3 and
P4 are orientably balanced as there exist γ-labelings of orientations of both paths
having value 0 (see Figure 3).
0
1
1
−1
1
0
0
0
−1
1
1
1
2
−1
3
2
−1
2
Figure 3: Illustrating γ-labelings of orientations of P2 , P3 , and P4
190
In fact, every path of order 3 or more is orientably balanced.
Proposition 3.2. For each integer n > 3, the path Pn is orientably balanced.
.
Let Pn : v1 , v2 , . . . , vn , let D be the orientation of Pn such that (v1 , v2 ) ∈
E(D) and (vi+1 , vi ) ∈ E(D) for 2 6 i 6 n − 1, and let f be any γ-labeling of D
for which f (v1 ) = 0, f (v2 ) = 1, and f (vn ) = 2. By Proposition 2.3(a), val(f ) =
(f (v2 ) − f (v1 )) + (f (v2 ) − f (vn )) = 1 + (−1) = 0. Thus Pn is orientably balanced.
Since P3 is orientably balanced, the star K1,2 is orientably balanced. The stars
K1,3 and K1,4 are also orientably balanced, as shown in Figure 4.
1
2
−3
−1
3
3
0
4
4
2
1
−2
0
−2
1
3
Figure 4: Illustrating γ-labelings of orientations of K1,3 and K1,4
Many other stars are orientably balanced.
Proposition 3.3. For each integer n > 3, every star K1,n−1 , where n 6≡ 2
(mod 4), is orientably balanced.
.
Let V (K1,n−1 ) = {v, v1 , v2 , . . . , vn−1 }, where v is the central vertex of
K1,n−1 . We consider two cases, according to whether n is odd or n is even.
1. n is odd. Then n = 2k + 1 for a positive integer k. Let D be the
orientation of K1,n−1 such that (vi , v) ∈ E(D) for 1 6 i 6 2k = n − 1. Define
the γ-labeling f of D by f (v) = k, f (vi ) = i − 1 for 1 6 i 6 k, and f (vi ) = i for
k + 1 6 i 6 2k. Figure 5 illustrates the labeling f of the orientation D of K1,6 for
k = 3.
v1
0
v2
1
3
2
v3
2
1
v4
4
−1
v5
v6
5
6
−2 −3
3
v
Figure 5: Illustrating the labeling f of the orientation D of K1,6 in Proposition 3.3 for
k = 3 and n = 7
191
Observe that
val(f ) =
2k
X
[f (v) − f (vi )] =
k
X
i+
k
X
i=1
i=1
=
[k − (i − 1)] +
i=1
k
X
2k
X
(k − i)
i=k+1
(−i) = 0.
i=1
Therefore, K1,n−1 is orientably balanced when n is odd.
2. n is even. Since n 6≡ 2 (mod 4), it follows that n ≡ 0 (mod 4) and so
n = 4k for some positive integer k. Let D 0 be the orientation of K1,n−1 such that
(v, vk+1 ) ∈ E(D0 ) and (vi , v) ∈ E(D0 ) for 1 6 i 6 4k − 1 = n − 1 and i 6= k + 1.
Furthermore, let g be the γ-labeling of D 0 defined by g(v) = 2k, g(vi ) = i − 1 for
1 6 i 6 2k, and g(vi ) = i for 2k + 1 6 i 6 4k − 1. Figure 6 illustrates the labeling g
of the orientation D 0 of K1,11 , where then k = 3 and n = 12.
v4
3
v3
v2
2
1
v1
0
5
4
−3
v5
4
2
v6
5
1
v7
7
v8
8
−1
−2
v9
v10
10
v11
−3
11
−4
9
−5
6
6
v
Figure 6: Illustrating the labeling g of the orientation D 0 of K1,11 in Proposition 3.3 for
k = 3 and n = 12
Observe that
val(g) = [g(vk+1 ) − g(v)] +
X
[g(v) − g(vi )]
16i64k−1
i6=k+1
X
2k−1
2k−1
2k X
X 2k−1
X
=
i − 2k +
(−i)
=
i +
(−i)
= 0.
i=1
i=1
i=1
i=1
Note that the −2k occurs because the arc (v, vk+1 , ) contributes −k to val(g) rather
than k. Therefore, K1,n−1 is orientably balanced for n is even and n ≡ 0 (mod 4).
An infinite class of orientably balanced graphs can be constructed from each orientably balanced graph. By a subdivision of a graph G, we mean a graph obtained
192
from G by inserting vertices (of degree 2) into one or more of the edges of G. An
elementary subdivision of a graph G is a graph obtained from G by inserting exactly
one vertex (of degree 2) into an edge of G.
Proposition 3.4. Every subdivision of an orientably balanced graph is orientably
balanced.
.
Let G be an orientably balanced graph and D an orientation of G. We
show, in fact, that if f is a γ-labeling of D and H is a subdivision of G, then there
exists an orientation F of H and a γ-labeling g of F such that val(g : F ) = val(f : D),
which implies that if f is a balanced γ-labeling of D, then g is a balanced γ-labeling
of F . It suffices to verify this statement for an elementary subdivision of G. Suppose
that H is the subdivision of G obtained by replacing the edge uv of G by the two
edges uw and wv. Let D be an orientation of G with (u, v) ∈ E(D) and let f be a
γ-labeling of D. Define an orientation F of H such that (u, w), (w, v) ∈ E(F ) and
(x, y) ∈ E(F ) if (x, y) ∈ E(D) and (x, y) 6= (u, v). Suppose that the size of G is m.
Then the size of H is m + 1. Define a γ-labeling g of F by g(x) = f (x) if x 6= w and
g(w) = m + 1. Then
X
X
g 0 (e) + g 0 ((u, w)) + g 0 ((w, v))
g 0 (e) =
val(g : F ) =
e∈E(F )−{(u,w),(w,v)}
e∈E(F )
=
=
=
X
e∈E(D)−{(u,v)}
X
e∈E(D)−{(u,v)}
X
f 0 (e) + [g(v) − g(u)]
0
f (e) + [f (v) − f (u)]
f 0 (e) = val(f : D),
e∈E(D)
producing the desired result.
By Proposition 3.4, all the graphs shown in Figure 7 are orientably balanced.
Figure 7: Orientably balanced graphs that are subdivisions of K1,3 and K1,4
We have already noted that the graph K2 = K1,1 is not orientably balanced.
Actually, the stars K1,5 and K1,9 also fail to be orientably balanced, as is the case
with the trees T1 and T2 in Figure 8.
193
K2
K1,5
K1,9
T1
T2
Figure 8: Trees that are not orientably balanced
Every vertex of the tree K1,3 has odd degree and K1,3 is orientably balanced. On
the other hand, every vertex in each tree shown in Figure 8 also has odd degree and
we have stated that none of these trees are orientably balanced. We now verify this
statement by showing that there are many trees that fail to be orientably balanced.
Proposition 3.5. Let T be a tree of order n > 2, every vertex of which has odd
degree. Then T is orientably balanced if and only if n ≡ 0 (mod 4).
.
Since every vertex of T has odd degree, n is even. Thus n ≡ 2 (mod 4)
or n ≡ 0 (mod 4). Assume that n ≡ 2 (mod 4). We show that T is not orientably
balanced. Let D be any orientation of T and let f be a γ-labeling of D. Then
val(f ) =
X
(f (v) − f (u)).
(u,v)∈E(D)
Observe that val(f ) is even if and only if
X
(f (v) + f (u)) =
X
(degT x)f (x)
x∈V (T )
uv∈E(T )
is even. Since T has order 4k + 2 for some nonnegative integer k, it follows that
2k + 1 vertices of T are assigned an even label and 2k + 1 vertices of T are assigned
an odd label. Because every vertex of T has odd degree,
X
(degT (x))f (x)
x∈V (T )
is odd. Thus, val(f ) is odd and so T is not orientably balanced.
Conversely, assume that n ≡ 0 (mod 4). We show that T is orientably balanced.
Then n = 4k for some positive integer k. We proceed by induction on k. If k = 1,
then T = K1,3 and so T is orientably balanced by Proposition 3.3. Assume that
every tree of order 4k, each of whose vertices has odd degree, is orientably balanced.
Now suppose that T is a tree of order 4k + 4 such that every vertex of T has odd
degree. If T is a star, then it follows by Proposition 3.3 that T is orientably balanced.
Thus, we may assume that T is not a star and so diam(T ) > 3.
194
Let u and v be two vertices of T such that d(u, v) = diam(T ). Thus u and v
are end-vertices of T . Suppose that u is adjacent to x in T and v is adjacent to
y in T . Then x 6= y, degT x > 3, and degT y > 3. Let u0 ∈ N (x) − {u} and
v 0 ∈ N (y) − {v} be end-vertices of T . Furthermore, let T 0 = T − {u, u0 , v, v 0 }. Then
T 0 is a tree of order 4k and every vertex of T 0 has odd degree. By the induction
hypothesis, T 0 is orientably balanced. Let D 0 be a balanced orientation of T 0 and
let f 0 : V (D0 ) → {0, 1, 2, . . . , 4k − 1} be a balanced γ-labeling of D 0 . Define an
orientation D of T with
E(D) = E(D0 ) ∪ {(u, x), (x, u0 ), (v, y), (y, v 0 )}.
Now define a γ-labeling f of D by

4k






 4k + 1

f (w) = 4k + 3




4k + 2



 0
f (w)
if w = u,
if w = u0 ,
if w = v,
if w = v 0 ,
if w ∈ V (D) − {u, u0 , v, v 0 }.
By Lemma 1.1 and Observation 2.1, it follows that
val(f : D) = val(f 0 : D0 ) + [(4k + 1) − (4k)] + [(4k + 2) − (4k + 3)] = 0.
Therefore, D is a balanced orientation of T and so T is orientably balanced.
As we have just seen, if T is a tree of order n, where n ≡ 2 (mod 4) and every
vertex of T has odd degree, then there exists no orientation of T for which there is a
γ-labeling f of D with val(f ) = 0. On the other hand, there is a related observation
we can make.
Proposition 3.6. If T is a tree of order n, where n ≡ 2 (mod 4) and every vertex
of T has odd degree, then there is an orientation D of T and a γ-labeling f of D
such that val(f : D) = 1.
.
Let n = 4k + 2 for some nonnegative integer k. Since the result
holds for n = 2, we may assume that k > 1. Let u and v be two vertices of T
such that d(u, v) = diam(T ). Necessarily, u and v are end-vertices of T . Suppose
that u is adjacent to x in T . Then degT x > 3. Then there exists an end-vertex
u0 ∈ N (x) − {u} of T . Then T 0 = T − {u, u0 } is a tree of order 4k and every vertex
of T 0 has odd degree. By Proposition 3.5, T 0 is orientably balanced and so there is
195
a balanced orientation D 0 of T 0 . Suppose that f 0 : V (D0 ) → {0, 1, 2, . . . , 4k − 1} be
a balanced γ-labeling of D 0 . Define an orientation D of T with
E(D) = E(D0 ) ∪ {(u, x), (x, u0 )}.
Now define a γ-labeling f of D by

n − 2 = 4k


f (w) = n − 1 = 4k + 1

 0
f (w)
if w = u,
if w = u0 ,
if w ∈ V (D) − {u, u0 }.
By Lemma 1.1 and Observation 2.1, it follows that
val(f : D) = val(f 0 : D0 ) + [(n − 1) − (n − 2)] = 1,
as desired.
The following is an immediate consequence of (1) and Proposition 3.6.
Corollary 3.7. If T is a tree of order n, where n ≡ 2 (mod 4) and every vertex
of T has odd degree, then there is an orientation D of T and a γ-labeling f of D
such that val(f : D) = −1.
We are now prepared to present a characterization of all trees that are orientably
balanced.
Theorem 3.8. A tree T of order n > 2 is orientably balanced if and only if either
n 6≡ 2 (mod 4) or T contains a vertex of even degree.
.
By Proposition 3.5, it suffices to show that if T is a tree of order n > 2
for which either n 6≡ 2 (mod 4) or T contains a vertex of even degree, then T is
orientably balanced. If every vertex of T has odd degree and n 6≡ 2 (mod 4), then
n ≡ 0 (mod 4). By Proposition 3.5 T is orientably balanced. Thus, we may assume
that at least one vertex of T has even degree.
We proceed by induction on the order n of T . It is easy to verify that the result
holds for 3 6 n 6 6. Assume that every tree of order n with 2 6 n < k (where k > 7)
having at least one vertex of even degree is orientably balanced. Let T be a tree of
order k having at least one vertex of even degree. If T is a path or T is a star, then
T is orientably balanced by Propositions 3.2 and 3.3. Thus, we may assume that
T 6= Pn and T 6= K1,n−1 .
First, suppose that T has exactly one vertex of degree 3 or more, say v ∈ V (T ) and
degT v = a > 3. Then T is a subdivision of the star K1,a . If a + 1 6≡ 2 (mod 4), then
196
K1,a is orientably balanced by Proposition 3.3 and so T is orientably balanced by
Proposition 3.4. Thus, we may assume that a + 1 ≡ 2 (mod 4). Then a = 4k + 1 > 5
for some positive integer k. Thus T contains at least five end-vertices. Let u be
an end-vertex of T such that d(u, v) is the maximum among all end-vertices of T .
Since T is not a star, d(u, v) > 2. Let x1 , x2 , y1 , y2 be end-vertices of T distinct
~ 1 and
from u, let Q1 be the x1 − x2 path in T and Q2 the y1 − y2 path in T . Let Q
~
~
Q2 be orientations of Q1 and Q2 , respectively, such that Q1 is a directed x1 − x2
~ 2 is a directed y1 − y2 path. Furthermore, let X = V (Q1 ) − {v} and
path and Q
Y = V (Q2 ) − {v}. Now let T 0 = T − (X ∪ Y ). Since T 0 contains the v − u path of
order at least 3, it follows that T 0 is a tree of order l > 3 having at least one vertex
of even degree. By the induction hypothesis, T 0 is orientably balanced. Let D 0 be
a balanced orientation of T 0 and let f 0 : V (D0 ) → {0, 1, 2, . . . , l − 1} be a balanced
γ-labeling of D0 . Define an orientation D of T with
~ 1 ) ∪ E(Q
~ 2 ).
E(D) = E(D0 ) ∪ E(Q
Now define a γ-labeling f of D such that f (w) = f 0 (w) if w ∈ V (D0 ) and f (x1 ) = l,
f (x2 ) = l + 1, f (y1 ) = l + 3, and f (y2 ) = l + 2. By Lemma 1.1 and Observation 2.1,
it follows that
val(f : D) = val(f 0 : D0 ) + [(l + 1) − (l)] + [(l + 2) − (l + 3)] = 0.
Thus T is orientably balanced.
Hence, we may assume that T has at least two vertices of degree 3 or more. Among
all vertices of degree 3 or more in T , let u and v be two such vertices for which d(u, v)
is maximum. Let u1 and u2 be end-vertices of T such that the only vertex of degree
3 or more on the u1 − u2 path Qu in T is u. Similarly, let v1 and v2 be end-vertices
of T such that the only vertex of degree 3 or more on the v1 − v2 path Qv in T is v.
Let
S1 = V (Qu ) − {u} and S2 = V (Qv ) − {v}.
Then |S1 ∪ S2 | > 4. Furthermore, if S1 ∪ S2 − {u1 , u2 , v1 , v2 } 6= ∅, then every vertex
in S1 ∪ S2 − {u1 , u2 , v1 , v2 } has degree 2. Let T 0 = T − (S1 ∪ S2 ). Then T 0 is a
~ u and Q
~ v be orientations of Qu and Qv ,
tree of order l = k − |S1 ∪ S2 | > 2. Let Q
~
~ v is a directed v1 − v2
respectively, such that Qu is a directed u1 − u2 path and Q
path. We consider two cases.
1. T 0 has a vertex of even degree or T 0 contains only vertices of odd
degree and l ≡ 0 (mod 4). By the induction hypothesis or by Proposition 3.5, T 0 is
orientably balanced. Let D 0 be a balanced orientation of T 0 and let f 0 : V (D0 ) →
{0, 1, 2, . . . , l − 1} be a balanced γ-labeling of D 0 . Define an orientation D of T with
~ u ) ∪ E(Q
~ v ).
E(D) = E(D0 ) ∪ E(Q
197
Now define a γ-labeling f of D such that f (w) = f 0 (w) if w ∈ V (D0 ) and f (u1 ) = l,
f (u2 ) = l + 1, f (v1 ) = l + 3, and f (v2 ) = l + 2. By Lemma 1.1 and Observation 2.1,
it follows that
val(f : D) = val(f 0 : D0 ) + [(l + 1) − (l)] + [(l + 2) − (l + 3)] = 0.
2. T 0 contains only vertices of odd degree and l ≡ 2 (mod 4). By
Proposition 3.6, there is an orientation D 0 of T 0 and a γ-labeling f 0 : V (D0 ) →
{0, 1, 2, . . . , l − 1} of D 0 such that val(f 0 : D0 ) = 1. In this case, S1 ∪ S2 contains at
least one vertex of degree 2 and so |S1 ∪ S2 | > 5. Thus the order of T is n > l + 5.
Again, define an orientation D of T with
~ u ) ∪ E(Q
~ v ).
E(D) = E(D0 ) ∪ E(Q
Now define a γ-labeling f of D such that f (w) = f 0 (w) if w ∈ V (D0 ) and f (u1 ) = l,
f (u2 ) = l + 1, f (v1 ) = l + 4, and f (v2 ) = l + 2. By Lemma 1.1 and Observation 2.1,
it follows that
val(f : D) = val(f 0 : D0 ) + [(l + 1) − (l)] + [(l + 2) − (l + 4)] = 1 + 1 + (−2) = 0.
In each case, there is a balanced orientation D of T and so T is orientably balanced.
Next, we determine all connected orientably balanced graphs. In order to do this,
we need an additional information. If we allow ourselves to work with “modified” γlabelings, then there is a result related to Proposition 3.6 that holds for all nontrivial
trees.
Proposition 3.9. If T is a tree of order n > 2, then there is an orientation D of
T and an injective function f : V (D) → {0, 1, 2, . . . , n} with val(f : D) = 1.
.
First, assume that T = Pn : v1 , v2 , . . . , vn . Let D be the orientation
of Pn that is a directed v1 − vn path and let f be any γ-labeling of D for which
f (v1 ) = 0, f (vn ) = 1. By Proposition 2.3(a), val(f ) = f (vn ) − f (v1 ) = 1.
Next, assume that T 6= Pn . Then T contains a vertex of degree 3 or more. Let w
be a vertex of degree 3 or more lying on a u − v path Q in T such that u and v are
end-vertices of T and for which every vertex in V (Q) − {w} has degree at most 2.
~ be an orientation
Let T 0 = T − (V (Q) − {w}). Then T 0 is a tree of order l > 2. Let Q
of Q that is a directed u − v path. We consider two cases.
1. T 0 is orientably balanced. Let D 0 be a balanced orientation of T 0 and let
0
0
f : V (D ) → {0, 1, 2, . . . , l − 1} be a balanced γ-labeling of D 0 . Define an orientation
D of T by
~
E(D) = E(D0 ) ∪ E(Q).
198
Now define a γ-labeling f of D such that f (x) = f 0 (x) if x ∈ V (D0 ), f (u) = l, and
f (v) = l + 1. By Lemma 1.1 and Observation 2.1, it follows that
val(f : D) = val(f 0 : D0 ) + [(l + 1) − (l)] = 0 + 1 = 1.
2. T 0 is not orientably balanced. By Theorem 3.8 and Corollary 3.7, there
is an orientation D 0 of T 0 and a γ-labeling f 0 : V (D0 ) → {0, 1, 2, . . . , l − 1} of D 0 such
that val(f 0 : D) = −1. Again, define an orientation D of T with
~
E(D) = E(D0 ) ∪ E(Q).
Since n > l + 2, we can define an injective function f : V (D) → {0, 1, 2, . . . , n} such
that f (x) = f 0 (x) if x ∈ V (D0 ), f (u) = l, and f (v) = l + 2. Then
val(f : D) = val(f 0 : D0 ) + [(l + 2) − (l)] = −1 + 2 = 1,
as desired.
Again, the following is an immediate consequence of (1) and Proposition 3.9.
Corollary 3.10. If T is a tree of order n > 2, then there is an orientation D of
T and an injective function f : V (D) → {0, 1, 2, . . . , n} with val(f : D) = −1.
With the aid of Theorem 3.8, Proposition 3.9, and Corollary 3.10, we can now
determine all connected orientably balanced graphs.
Theorem 3.11. A connected graph G of order n > 2 is orientably balanced
unless G is a tree, n ≡ 2 (mod 4), and every vertex of G has odd degree.
.
By Theorem 3.8, we may assume that G is connected graph of order
n > 2 that is not a tree. Since G is not a tree, G0 = G contains at least one cycle.
Let C1 be a cycle in G and let G1 = G0 − E(C1 ). If G1 is a forest, then let F = G1 .
If G1 is not a forest, then G1 contains a cycle C2 . Let G2 = G1 − E(C2 ). If G2 is
a forest, then let F = G2 . Otherwise, we continue this procedure until a spanning
forest F = Gk−1 − E(Ck ) of G is produced, where k > 1. If all components of F
are trivial, then G is Eulerian and so G is orientably balanced by Proposition 3.1.
Hence we may assume that F has p > 1 nontrivial components T1 , T2 , . . . , Tp , where
each tree Ti has order ni > 2 for 1 6 i 6 p. Then the order of G is
n > n1 + n2 + . . . + n p .
199
Since G is not a tree, the size m of G is at least n. For each cycle Cj removed from
~ j be an orientation of Cj that is a directed cycle. We
G, where 1 6 j 6 k, let C
consider two cases.
1. Every tree Ti (1 6 i 6 p) is orientably balanced. For each integer
i with 1 6 i 6 p, let Di be a balanced orientation of Ti and let fi : V (Di ) →
{0, 1, 2, . . . , ni − 1} be a balanced γ-labeling of Di . Define an orientation D of G
with
(3)
E(D) =
p
[
i=1
E(Di ) ∪
k
[
j=1
~j ) .
E(C
Now define a γ-labeling g : V (D) → {0, 1, 2, . . . , m} of D such that
(4)

f1 (v)





f2 (v) + n1



g(v) = f3 (v) + n1 + n2


..



.



fp (v) + n1 + n2 + . . . + np−1
if v ∈ V (T1 ),
if v ∈ V (T2 ),
if v ∈ V (T3 ),
..
.
if v ∈ V (Tp ).
Observe that the restriction g1 of g to D1 is f1 and, for 2 6 i 6 p, the restriction
gi of g to Di defined by gi (v) = fi (v) + n1 + n2 + . . . + ni−1 for v ∈ V (Di ). Thus
~j
val(gi : Di ) = val(fi : Di ) = 0 for 1 6 i 6 p. Furthermore, each directed cycle C
0
~
is Eulerian for 1 6 j 6 k and so the restriction gj of g to each Cj is balanced. By
Lemma 1.1,
val(f, D) =
X
p
i=1
val(fi : Di ) +
X
k
j=1
val(gj0
~j )
:C
= 0.
2. Some tree Ti (1 6 i 6 p) is not orientably balanced. Assume, without
loss of generality, that T1 , T2 , . . . , Ta are not orientably balanced, where 1 6 a 6 p,
and Ta+1 , Ta+2 , . . . , Tp are orientably balanced if a < p. By Theorem 3.8, for each
integer i with 1 6 i 6 a, Ti is a tree of order ni where ni ≡ 2 (mod 4) and every
vertex of Ti has odd degree. We consider two subcases, according to whether a is
even or a is odd.
2.1. a is even. Then a = 2b for some positive integer b. For each
integer i with 1 6 i 6 p, let Di be an orientation of Ti and let fi : V (Di ) →
{0, 1, 2, . . . , ni − 1} be a γ-labeling of Di such that (i) val(fi : Di ) = 1 for 1 6 i 6 b,
(ii) val(fi : Di ) = −1 for b + 1 6 i 6 a, and (iii) val(fi : Di ) = 0 for any integers i
with a+1 6 i 6 p. Define the orientation D of G as described in (3) and a γ-labeling
200
g of D that satisfies (4). An argument similar to the one used in Case 1 shows that
g is a balanced γ-labeling of D.
2.2. a is odd. Then a = 2b+1 for some nonnegative integer b. Assume,
without loss of generality, that n1 > ni for every integer i with 1 6 i 6 a, that is, T1
has the largest order among all trees Ti for 1 6 i 6 a. We consider two subcases.
2.2.1. n1 > 3. Thus n1 > 6 and T1 contains at least one vertex of odd
degree 3 or more by Theorem 3.8. Let u be a vertex of degree 3 or more such that
there exists a u1 − u2 path Qu in T1 for which u1 and u2 are end-vertices of T1 and
u is the only vertex on Qu whose degree is 3 or more. Since T1 contains no vertex
~ u be an orientation of Qu that is a
of even degree, Qu is a path of length 2. Let Q
0
u1 − u2 directed path. Let T = T1 − (V (Qu ) − {u}) = T1 − {u1 , u2 }. Then T 0 is
a tree of order l = n1 − 2. By Proposition 3.9, there is an orientation D 0 of T 0 and
an injective function f 0 : V (D0 ) → {0, 1, 2, . . . , l} with val(f 0 : D0 ) = −1. Define an
orientation D1 of T1 with
~ u ).
E(D1 ) = E(D0 ) ∪ E(Q
Since n1 = l + 2, we can define an injective function f1 : V (D1 ) → {0, 1, . . . , l + 2}
such that f1 (w) = f 0 (w) if w ∈ V (D0 ) and f1 (u1 ) = l + 1, and f (u2 ) = l + 2. By
Lemma 1.1 and Observation 2.1, it follows that
val(f1 : D1 ) = val(f 0 : D0 ) + [(l + 2) − (l + 1)] = (−1) + 1 = 0.
For each integer i with 2 6 i 6 p, let Di be an orientation of Ti and let fi : V (Di ) →
{0, 1, 2, . . . , ni −1} be a γ-labeling of Di such that (i) val(fi : Di ) = 1 for 2 6 i 6 b+1,
(ii) val(fi : Di ) = −1 for b + 2 6 i 6 a, and (iii) val(fi : Di ) = 0 for a + 1 6 i 6 p.
Define an orientation D of G as described in (3) and a γ-labeling g of D such that

f1 (v)






 f2 (v) + n1 + 1

g(v) = f3 (v) + n1 + n2 + 1


..



.



fp (v) + n1 + n2 + . . . + np−1 + 1
if v ∈ V (T1 ),
if v ∈ V (T2 ),
if v ∈ V (T3 ),
..
.
if v ∈ V (Tp ).
An argument similar to the one used in Case 1 shows that g is a balanced γ-labeling
of D.
2.2.2. n1 = 2. Thus Ti = K2 for all integers i with 1 6 i 6 a.
Suppose first that a > 3. For each integer i with 1 6 i 6 3, let V (Ti ) = {ui , vi }
and let Di be a directed ui − vi path whose single arc is (ui , vi ). Define an injective
201
function f 0 : V (D1 ) ∪ V (D2 ) ∪ V (D3 ) → {0, 1, . . . , 6} by f 0 (u1 ) = 0, f 0 (v1 ) = 1,
f 0 (u2 ) = 2, f 0 (v2 ) = 3, f 0 (u3 ) = 6, and f 0 (v3 ) = 4. Let D0 = D1 ∪ D2 ∪ D3 .
Then val(f 0 : D0 ) = 0. For each integer i with 4 6 i 6 p, let Di be an orientation
of Ti and let fi : V (Di ) → {0, 1, 2, . . . , ni − 1} be a γ-labeling of Di such that (i)
val(fi : Di ) = 1 for 4 6 i 6 b + 2, (ii) val(fi : Di ) = −1 for b + 3 6 i 6 a, and (iii)
val(fi : Di ) = 0 for a + 1 6 i 6 p. Define an orientation D of G as described in (3)
and a γ-labeling g of D such that
 0
f (v)





f4 (v) + 7



g(v) = f5 (v) + n4 + 7


 ..


.



fp (v) + n4 + n5 + . . . + np−1 + 7
if v ∈ V (D0 ),
if v ∈ V (T4 ),
if v ∈ V (T5 ),
..
.
if v ∈ V (Tp ).
An argument similar to the one used in Case 1 shows that g is a balanced γ-labeling
of D.
Hence we may assume that a = 1. Let T1 : u, v and let D1 be a u − v directed
path whose single arc is (u, v). We consider two subcases.
2.2.2.1. p > 2. Define f1 on V (D1 ) by f (u) = 1 and f (v) = 0.
Thus val(f1 : D1 ) = −1. Let D2 be an orientation of T2 and let f2 : V (D2 ) →
{0, 1, 2, . . . , n2 } be an injective function such that val(f2 : D2 ) = 1. For each integer
i with 3 6 i 6 p, let Di be a balanced orientation of Ti and let fi : V (Di ) →
{0, 1, 2, . . . , ni − 1} be a balanced γ-labeling of Di . Define an orientation D of G as
described in (3) and a γ-labeling g of D such that

f1 (v)






 f2 (v) + 2

g(v) = f3 (v) + n2 + 2


..



.



fp (v) + n2 + . . . + np−1 + 2
if v ∈ V (T1 ),
if v ∈ V (T2 ),
if v ∈ V (T3 ),
..
.
if v ∈ V (Tp ).
Then g is a balanced γ-labeling of D.
2.2.2.2. p = 1. The cycle C1 may contain both, exactly one of, or
neither of u and v but surely C1 does not contain the edge uv. Furthermore, if C1
contains both u and v, then the order of C1 is 4 or more. In any case, C1 contains
at least two vertices of V (G) − {u, v}. Let x, y ∈ V (C1 ) − {u, v}.
Consider any injective function g : V (G) → {0, 1, . . . , n} such that g(u) = 2,
g(v) = 0, g(x) = 3, and g(y) = 4. Now let S1 be the orientation of C1 that creates
202
two directed x − y paths. Define an orientation D of G by
E(D) = E(D1 ) ∪ E(S1 ) ∪
[
k
j=2
~
E(Cj ) .
Thus g is a γ-labeling on D and
val(g : D) = val(g : D1 ) + val(g : S1 ) +
k
X
~ j ) = −2 + 2 + 0 = 0.
val(g : C
j=2
Hence g is a balanced γ-labeling of D.
This completes the proof.
References
[1] G. Chartrand, D. Erwin, D. W. VanderJagt, P. Zhang: γ-labelings of graphs. Bull. Inst.
Combin. Appl. 44 (2005), 51–68.
[2] G. Chartrand, D. Erwin, D. W. VanderJagt, P. Zhang: On γ-labelings of trees. Discuss.
Math., Graph Theory 25 (2005), 363–383.
[3] G. Chartrand, P. Zhang: Introduction to Graph Theory. McGraw-Hill, Boston, 2005.
[4] J. A. Gallian,: A dynamic survey of graph labeling. Electron. J. Combin. 5 (1998), Dynamic Survey 6, pp. 43.
[5] S. W. Golomb: How to number a graph. Graph Theory Comp. Academic Press, New
York, 1972, pp. 23–37.
[6] A. Rosa: On certain valuations of the vertices of a graph. Theory Graphs, Proc. Int.
Symp. Rome 1966, Gordon and Breach, New York, 1967, pp. 349–355.
Authors’ addresses: Futaba Okamoto, University of Wisconsin-La Crosse, Mathematics
Department, La Crosse, WI 5461, USA; Ping Zhang, Western Michigan University, Department of Mathematics, Kalamazoo, MI 49008, USA, e-mail: ping.zhang@wmich.edu; Varaporn Saenpholphat, Srinakharinwirot University, Department of Mathematics, Sukhumvit
Soi 23, Bangkok, 10110, Thailand.
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